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3 years ago

Two stars that are in the same constellation:

Physics

1 answer:

sammy [17]3 years ago

7 0

Answer:

A binary star is a star system consisting of two stars orbiting around their common barycenter.

<h3>❣️(◍Jess bregoli◍)❣️</h3>

#keep learning

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Sequence eye light enters the eye

Burka [1]

Answer:

Well it is the courtnes

Explanation:

8 0

3 years ago

When none of the magnetic domains in a material will stay aligned, the material is called __________________?

Kisachek [45]

The material is called paramagnetic or non magnetic material. Opposite to this are materials whose magnetic domains are aligned and possess asmagnetism which is due to its ferrous or iron content.

7 0

3 years ago

A uniform rod 8m long weighing 5kg is supported horizontally by two vertical parallel strings at p and q and at distance 2m and

Gala2k [10]

Answer:

Fp = 36 N

Fq = 58 N

Explanation:

Let the left end be the reference end with string p closest to it.

Let CCW moments be positive

Sum moments about p to zero

1(9.8)[2 - 1] + Fq[6 - 2] - 5(9.8)[8/2 - 2] - 1.5(9.8)[5 - 2] - 2(9.8)[7 - 2] = 0

Fq[4] = 23.5(9.8)

Fq = 57.575 ≈ 58 N

Sum moments about q to zero

1(9.8)[6 - 1] - Fp[6 - 2] + 5(9.8)[6 - 8/2] + 1.5(9.8)[6 - 5] - 2(9.8)[7 - 6] = 0

Fp = 35.525 N

Sum vertical forces to zero

Fp + 57.575 - (9.8)(1 + 1.5 + 2 + 5) = 0

Fp = 35.525 ≈ 36 N

4 0

2 years ago

You run completely around a 400m track in 80s. What was your average velocity?

dalvyx [7]

Answer:

V=?

S=400m

t=80s

V=S/t

V=400/80

V=5m/sec

6 0

3 years ago

PLEASE HELP

Fed [463]

Answer:

a) t =12[s]; b) x = 348[m]

Explanation:

We can solve this problem using the following kinematics equations:

$v_{f}= v_{o}+a*t$

where:

vf = final velocity = 12 [m/s]

vo= initial velocity = 6 [m/s]

a = acceleration = 0.5[m/s^2]

t = time [s]

Now clearing the time t, we have:

$t=\frac{v_{f} -v_{o} }{a} \\t = \frac{12-6}{0.5} \\t=12[s]$

We can calculate the displacement for the first 12 [s] then using the equation for the constant velocity we can calculate the other displacement for the 20[s].

$v_{f}^{2}= v_{o}^{2}+2*a*x_{1} \\therefore\\x_{1} =\frac{v_{f}^{2}-v_{o}^{2}}{2*a} \\x_{1} =\frac{12^{2}-6^{2}}{2*.5}\\x_{1} =108[m]$

The we can calculate the second displacement for the constant velocity:

$x_{2} =x_{o}+v*t_{2} \\ x_{2} =0+12*(20)\\x_{2} =240[m]$

x = x1 + x2

x = 108 + 240

x = 348[m]

5 0

4 years ago